Question

Combining 0.296 mol of Fe_2O_3 with excess carbon produced 13.4 g of Fe. Fe_2O_3 + 3C rightarrow 2Fe + 3CO What is the actual yield of iron in moles? What was the theoretical yield of iron in moles? What was the percent yield?

Answer 1

Fe2o3 + 3C ---> 2Fe + 3co

No of mole of Fe2O3 = 0.296 mole

from equation, 1 mol Fe2o3 = 3 mol C = 2 mole Fe

as C-excess , limiting reagent = Fe2o3

actual yield in moles = 13.4/55.845 = 0.24 mole

Theoretical yield in moles = 0.296*2 = 0.592 mole

% yield = actual yield / Theoretical yield *100

         = 0.24/0.592*100

         = 40.54%